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What is the equilibrium constant for the reaction between ethanoic acid and water, given that the dissociation constant (Ka) for ethanoic acid is 1.8 x 10^-5 at 25°C?

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The equilibrium constant for the reaction between ethanoic acid  CH3COOH  and water  H2O  can be determined using the dissociation constant  Ka  for ethanoic acid. The reaction can be written as:CH3COOH + H2O  CH3COO- + H3O+The dissociation constant  Ka  is given as 1.8 x 10^-5 at 25C. In this case, the Ka value is the equilibrium constant for the reaction, as it represents the ratio of the concentrations of the products to the reactants at equilibrium. Therefore, the equilibrium constant  K  for the reaction between ethanoic acid and water is:K = Ka = 1.8 x 10^-5

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